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Math Help - Big Oh in arithmetic expression

  1. #1
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    Big Oh in arithmetic expression

    Hello to everyone,

    Could somebody please tell me how do we solve exercises like the one I describe below:

    exp(1+(O(1/n))^2) = e + O(1/n)


    Thanks a lot,
    Tasos
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  2. #2
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    Hi Tasos:

    I see the equation, but I do not know what the instructions are for this exercise.

    Are exp(1) and e both the same constant in this equation?

    What are we to do with this equation?

    ~ Mark

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  3. #3
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    First of all, I want to thank you mmmbot for answering.

    Sorry if I didn't write it well. By writing exp I meant e to the power.

    So I mean: [e to the power of (1 + (O(1/n))^2)] = e + O(1/n)


    P.S.1: Could you tell me how to write it in the post so as it will appear correctly?
    P.S.2: What you say is right: exp(1) and e are the same thing. That is what I mean.
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  4. #4
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    I read it as follows.

    e^[1 + (O/n)^2] = e + O/n
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  5. #5
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    Quote Originally Posted by mmm4444bot View Post
    I read it as follows.

    e^[1 + (O/n)^2] = e + O/n

    Sorry, then. I didn't write O/n, but O(1/n).

    Is it the same thing?
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  6. #6
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    I'm checking to see whether or not O(1/n) is function notation ... be right back.

    A big "Oh, please excuse me".

    I thought O and n were constants. (I just saw the big Oh function for the first time at Wikipedia.)

    No, you are correct. O(1/n) is good.

    I cannot help you with this exercise.

    Now, I read it as follows.

    e^[1 + O(1/n)^2] = e + O(1/n)


    Again, sorry for goofing up your thread.

    ~ Mark
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  7. #7
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    Quote Originally Posted by mmm4444bot View Post
    I'm checking to see whether or not O(1/n) is function notation ... be right back.

    A big "Oh, please excuse me".

    I thought O and n were constants. (I just saw the big Oh function for the first time at Wikipedia.)

    No, you are correct. O(1/n) is good.

    I cannot help you with this exercise.

    Now, I read it as follows.

    e^[1 + O(1/n)^2] = e + O(1/n)


    Again, sorry for goofing up your thread.

    ~ Mark
    No problem, Mark. Don't worry.
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  8. #8
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    Quote Originally Posted by tasos View Post
    Hello to everyone,

    Could somebody please tell me how do we solve exercises like the one I describe below:

    exp(1+(O(1/n))^2) = e + O(1/n)


    Thanks a lot,
    Tasos
    Are you sure you put the parentheses at the right place? Because this is correct, but not optimal, as you shall see:

    I'll apply usual properties of the "big O" notation. If you don't know them, just ask, the proofs are very short.
    First you have \left(O\left(\frac{1}{n}\right)\right)^2=O\left(\f  rac{1}{n^2}\right), and it is more usual to write it this way.
    Then e^{1+O\left(\frac{1}{n^2}\right)}=e e^{O\left(\frac{1}{n^2}\right)}= e\left( 1+O\left(\frac{1}{n^2}\right)\right) =e + e\, O\left(\frac{1}{n^2}\right)=e+O\left(\frac{1}{n^2}  \right).
    I used the following expansion of the exponential at 0: e^{O(u)}=1+O(u) when u tends to 0, composed with the sequence \left(\frac{1}{n^2}\right)_n which tends to 0.

    Now, 0\leq\frac{1}{n^2}\leq \frac{1}{n}, hence the previous big O can be replaced by O\left(\frac{1}{n}\right), and this is what you need.

    Let us now suppose that the question was e^{\left(1+O\left(\frac{1}{n}\right)\right)^2}=1+O  \left(\frac{1}{n}\right), which seems more plausible to me.
    Then we would have: \left(1+O\left(\frac{1}{n}\right)\right)^2=1+O\lef  t(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)=1  +O\left(\frac{1}{n}\right) (as before, the second big O can be included in the first one), and by the same computation as above we conclude e^{\left(1+O\left(\frac{1}{n}\right)\right)^2}=e+O  \left(\frac{1}{n}\right).

    ps: about the way to write symbols, there's a section in the forum related to "LaTeX", and this is where you should look for help about this.
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  9. #9
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    Thank you very much Laurent for your answer. It is very thorough.

    However, I have a little question. How do you go from the left part to the right one in the following expression:

    Quote Originally Posted by Laurent View Post
    e e^{O\left(\frac{1}{n^2}\right)}= e\left( 1+O\left(\frac{1}{n^2}\right)\right)

    Thanks
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  10. #10
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    Quote Originally Posted by tasos View Post
    Thank you very much Laurent for your answer. It is very thorough.

    However, I have a little question. How do you go from the left part to the right one in the following expression:
    Thanks
    This is because of this result:

    Quote Originally Posted by Laurent View Post
    I used the following expansion of the exponential at 0: e^{O(u)}=1+O(u) when u tends to 0, composed with the sequence \left(\frac{1}{n^2}\right)_n which tends to 0.
    In fact, as soon as a function f is differentiable at 0, we have f(x)=f(0)+O(x) as x tends to 0 (because \frac{f(x)-f(0)}{x}\to f'(0) so that \frac{f(x)-f(0)}{x}=O(1) as x\to0).

    And if g(u)=O(u) as u\to0, then g(u)\to_{u\to 0} 0, so that we can compose: f(g(u))=f(0)+O(g(u))=f(0)+O(u). This can be written f(O(u))=f(0)+O(u) as u tends to 0.

    Now, in this expansion, you can replace u by any sequence which converges to 0. For instance, f(O\left(\frac{1}{n^2}\right))=f(0)+O\left(\frac{1  }{n^2}\right). If f=\exp, you get what I wrote and used.

    (I'm thinking of something that may have been confusing: when I wrote e\left( 1+O\left(\frac{1}{n^2}\right)\right), it meant e\times \left( 1+O\left(\frac{1}{n^2}\right)\right), not exponential of the parenthesis)
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  11. #11
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    Laurent, I want to thank you very much for your answers. They were very comprehensible.
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